24 divides a(a^2 - 1) when a is Odd

Theorem

Let $a \in \Z$ be an odd integer.

Then:

$24 \divides a \paren {a^2 - 1}$

where $\divides$ denotes divisibility.

Proof

First suppose that $a$ is not divisible by $3$.

Then from Square Modulo 24 of Odd Integer Not Divisible by 3:

$24 \divides \paren {a^2 - 1}$

from which the result follows immediately.

$\Box$

Now suppose that $3 \divides a$.

Then immediately:

$3 \divides a \paren {a^2 - 1}$

From Odd Square Modulo 8:

$8 \divides \paren {a^2 - 1}$

and so:

$8 \divides a \paren {a^2 - 1}$

We have from Coprime Integers: $3$ and $8$ that:

$3 \perp 8$

where $\perp$ denotes coprimality.

As we have that:

$8 \divides a \paren {a^2 - 1}$

and:

$3 \divides a \paren {a^2 - 1}$

it follows from Product of Coprime Factors that:

$24 \divides a \paren {a^2 - 1}$

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $15 \ \text {(a)}$